Question: Divide the following complex numbers. $ \dfrac{-4+5i}{-5-4i}$
Answer: We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${-5+4i}$ $ \dfrac{-4+5i}{-5-4i} = \dfrac{-4+5i}{-5-4i} \cdot \dfrac{{-5+4i}}{{-5+4i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-4+5i) \cdot (-5+4i)} {(-5-4i) \cdot (-5+4i)} = \dfrac{(-4+5i) \cdot (-5+4i)} {(-5)^2 - (-4i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-4+5i) \cdot (-5+4i)} {(-5)^2 - (-4i)^2} = $ $ \dfrac{(-4+5i) \cdot (-5+4i)} {25 + 16} = $ $ \dfrac{(-4+5i) \cdot (-5+4i)} {41} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-4+5i}) \cdot ({-5+4i})} {41} = $ $ \dfrac{{-4} \cdot {(-5)} + {5} \cdot {(-5) i} + {-4} \cdot {4 i} + {5} \cdot {4 i^2}} {41} $ Evaluate each product of two numbers. $ \dfrac{20 - 25i - 16i + 20 i^2} {41} $ Finally, simplify the fraction. $ \dfrac{20 - 25i - 16i - 20} {41} = \dfrac{0 - 41i} {41} = -i $